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Homework Help: Finding the equation of the intersection

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Plane 1:2x-3y+8z=7
    Plane 2:x-8y+z=14
    Plane 3:5x-14y+17z=28

    2. Relevant equations

    3. The attempt at a solution
    I took plane 1 and subtracted (plane 2)x2 to get
    13y+6z=-21 (I will refer to this as equation 1)

    Then I took (plane 2)x5 and subtracted plane 3 to get
    which simplifies to
    13y+6z=-16 (I will refer to this as equation 2)

    Then I was stumped because the two equations contradict each other and state 0=-5
    which leads me to believe that there is no values of x,y,z that can satisfy all planes

    although the answers in the book say
    "consistent [14/13,-21/13,0] + t[-61/13,-6/13,1]"

    I understand how they get the answer I just don't understand why, if equation 1 and 2 do not provide a valid statement why do they continue and find the parametric equations

    like let z = t
    13y=-6t -21
    y=-6/13t -21/13
    then substitute y into plane 2 to get
    x-8(-6/13z - 21/13) + z=14 and simplifies
    x=61/13z +14/13
    then z=0+t
    which gets the correct parametric equations which I tested by using different values of t and the points do satisfy all planes
    I guess the real question I have is why do they chose "equation 1" instead of "equation 2" which would yield a different result if you were to use that to fidn the parametric equations, which I'm assuming the parametric equations don't work and why do you continue after the elimination of equation 1 and 2 which results in 0=-5.
  2. jcsd
  3. Mar 30, 2012 #2


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    Hello cdot9. Welcome to PF !

    5×14 - 28 = 42, not 32.
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